利用格林函数求解二维拉普拉斯方程：第一象限格林函数推导

Alright, let's break down how to derive and verify this Green's function for the Poisson equation in the first quadrant—this is a classic method of images problem, so it makes sense once we map out the required image sources and check each condition step by step.

We're tasked with finding a Green's function (G(\mathbf{x}, \mathbf{\xi})) in the first quadrant (D: {(x,y) : x>0, y>0}) where:

• (\mathbf{x} = (x,y)) and (\mathbf{\xi} = (\xi_x, \xi_y)) are points in (D)
• The function satisfies the Poisson equation: (\nabla^2G = \delta(\mathbf{x} - \mathbf{\xi}))
• Boundary conditions:
  • Neumann boundary at (x=0): (\frac{\partial G}{\partial x}(0, y, \mathbf{\xi}) = 0) for (y>0)
  • Dirichlet boundary at (y=0): (G(x, 0, \mathbf{\xi}) = 0) for (x>0)

The method of images works by introducing "mirror" sources outside our domain to satisfy the boundary conditions. Here's what we need:

1. For the Dirichlet boundary (y=0): To enforce (G=0) at (y=0), we add a negative mirror source at ((\xi_x, -\xi_y)) (below the x-axis). This cancels out the potential from the original source at (y=0).
2. For the Neumann boundary (x=0): To enforce (\frac{\partial G}{\partial x}=0) at (x=0), we need to reflect both the original source and the y-axis mirror source across the y-axis. These are positive mirror sources at ((-\xi_x, \xi_y)) and ((-\xi_x, -\xi_y)) (left of the y-axis), since Neumann boundaries require a "reflection" rather than cancellation of the potential's derivative.

The fundamental 2D Green's function for a point source is (\frac{1}{4\pi}\ln\left(\frac{1}{|\mathbf{x} - \mathbf{\xi}|}\right)). Adding/subtracting the Green's functions for our four sources (original + 3 mirrors) gives us the total Green's function:
[
G = \frac{1}{4\pi}\ln\left(\frac{|\mathbf{x} - \mathbf{\xi}| \cdot |\mathbf{x} - (-\xi_x, \xi_y)|}{|\mathbf{x} - (\xi_x, -\xi_y)| \cdot |\mathbf{x} - (-\xi_x, -\xi_y)|}\right)
]
Substituting (|\mathbf{x} - \mathbf{a}|^2 = (x - a_x)^2 + (y - a_y)^2) (since (\ln(a^2) = 2\ln a), the square doesn't change the log's ratio), we get the given expression:
[
G = \frac{1}{4 \pi}\ln \left| \frac{((x - \xi_x)^2 + (y- \xi_y)^2)((x + \xi_x)^2 + (y- \xi_y)^2)}{((x - \xi_x)^2 + (y+ \xi_y)^2)((x + \xi_x)^2 + (y+ \xi_y)^2)} \right|
]

3.1 Dirichlet Boundary (y=0) #

Substitute (y=0) into the expression:

• Numerator: (((x-\xi_x)^2 + \xi_y^2)((x+\xi_x)2 + \xi_y^2))
• Denominator: (((x-\xi_x)^2 + \xi_y^2)((x+\xi_x)2 + \xi_y^2))

The numerator equals the denominator, so (\ln(1) = 0), which satisfies (G(x,0,\mathbf{\xi})=0). Perfect.

3.2 Neumann Boundary (x=0) #

First compute the partial derivative of (G) with respect to (x):
[
\frac{\partial G}{\partial x} = \frac{1}{4\pi} \left[ \frac{2(x-\xi_x)}{(x-\xi_x)^2+(y-\xi_y)2} + \frac{2(x+\xi_x)}{(x+\xi_x)^2+(y-\xi_y)2} - \frac{2(x-\xi_x)}{(x-\xi_x)^2+(y+\xi_y)2} - \frac{2(x+\xi_x)}{(x+\xi_x)^2+(y+\xi_y)2} \right]
]
Substitute (x=0):
[
\frac{\partial G}{\partial x}(0,y,\mathbf{\xi}) = \frac{1}{4\pi} \left[ \frac{-2\xi_x}{\xi_x^2+(y-\xi_y)2} + \frac{2\xi_x}{\xi_x^2+(y-\xi_y)2} - \frac{-2\xi_x}{\xi_x^2+(y+\xi_y)2} - \frac{2\xi_x}{\xi_x^2+(y+\xi_y)2} \right]
]
Each pair of terms cancels out to 0, so the partial derivative is 0—exactly what the Neumann boundary requires.

Each fundamental Green's function satisfies (\nabla^2G_i = \delta(\mathbf{x} - \mathbf{\xi}_i)), where (\mathbf{\xi}_i) is the source point. For our total Green's function:
[
\nabla^2G = \delta(\mathbf{x}-\mathbf{\xi}) + \delta(\mathbf{x}-(-\xi_x,\xi_y)) - \delta(\mathbf{x}-(\xi_x,-\xi_y)) - \delta(\mathbf{x}-(-\xi_x,-\xi_y))
]
In the first quadrant (D) (where (x>0,y>0)), the three mirror sources lie outside (D), so their delta functions are 0. Only the original source's delta function remains, so (\nabla^2G = \delta(\mathbf{x}-\mathbf{\xi})), which satisfies the Poisson equation.

内容的提问来源于stack exchange，提问作者Unicorn